Найти действительные x и y из условия равенства (1 + i)x + (2+i)y=3i+1

(1 + i)x + (2+i)y=3i+1

(i +1)x+(i+2)y=3i+1

xi+x×1+yi+y×2=3i+1

ix+yi+x+y2=3i+1

ix+x+2y=-yi+3i+1

ix+x=-yi+3i-2y+1

x(i+1)=-yi+3i-2y+1

x=-yi+3i-2y+1/i+1

(1 + i)x + (2+i)y=3i+1

(i+2)y+(i+1)x=3i+1

yi+y×2+xi+x×1=3i+1

yi+xi+2y+x=3i+1

iy+2y+x=-xi+3i+1

iy+2y=-xi+3i-x+1

y(i+2)=-xi+3i-x+1

y=-xi+3i-x+1/i+2